Friday, November 13, 2009

Fox 183


5 comments:

  1. Assume that the origin (0,0) is the centre of the smallest circle and that the centre of the semicircle is (d,0); hence its radius r = d+1
    Now it can be easily verified that the semicrcle and the smallest circle x^2+y^2=1 touch each other at (-1,0) as depicted. Our assumption that the centre of the semicircle lies on the x-axis seems, therefore, to be correct. Now let the semicircle intesect x^2+y^2 =4 in (x1,y1) in the first or second quadrant and let it intersect x^2+y^2=9 in the fourth quadrant in (x2,y2). We can easily show that x1=(4-d)/d while x2=(3-2d)/2d. Further, (x1+x2)=2d since the midpoint of (x1,y1) and (x2,y2) is (d,0). Thus,(4-d)/d + (3-2d)/2d =2d which gives d = V3 -1/2 or r = d+1 = V3 + 1/2 (Option D)
    Would that be correct?
    Ajit

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  2. Erratum: That's Option E
    Ajit

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  3. http://i38.tinypic.com/zvu244.jpg

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  4. Both answers are similar and correct.
    We actually had a little shorter solution like:
    cos(a)=-cos(180-a)
    which immediately gives a quadratic eqn leading to the solution, but we've lost it. And for some reason I just can't see it!!!
    Anyways...
    Thanks you both for the solution.

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